Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 43 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


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Problem

Show that the equation x^3 + 7x - 14(n^2 +1) = 0 has no integral root for any integer n.

Solution:

We note that 14(n^2 +1) - 7x = x^3 implies x^3 is divisible by 7. This implies x is divisible by 7 (as 7 is a prime number). Suppose x= 7x’. Hence we can rewrite the given equation as:

7^3 x'^3 + 7 \times 7 x' - 14 (n^2 +1 ) = 0 .

Cancelling out a 7 we have 7^2 {x'}^3 + 7{x'} = 2(n^2 +1) . Since 7 divides left hand side, it must also divide the right hand side. Since 7 cannot divide 2, it must divide n^2 + 1 as 7 and 2 are coprime. Note that 7 cannot divide n^2 +1 as square of a number always gives remainder 0, 1, 4, 2 when divided by 7 and never 6. But if n^2 + 1 is divisible by 7 then n^2 must give remainder 6 when divided by 7. Hence contradiction.

Necessary Lemma: square of a number always gives remainder 0, 1, 4, 2 when divided by 7

n \equiv 0, \pm 1 , \pm 2 , \pm 3 \mod 7\Rightarrow n^2 \equiv 0, 1, 4, 9 (=2) \mod 7

Key Ideas: Modular Arithmetic